Quadratic Equations form a fundamental part of competitive exams, testing candidates' mathematical reasoning and problemsolving abilities. These equations often serve as a bridge between basic algebraic concepts and more advanced mathematical applications, making them a crucial foundation for various topics. When preparing for Quadratic Equations, focus on fundamental concepts. Remember that quadratic equations have two roots, and these roots can be either real or imaginary. Having a good grasp of these basic concepts will help you solve questions more efficiently.
The equation is a valid expression with an equal sign (=), representing the balancing of the given terms on both sides of it. Solving these equations, we may get some root values of the unknown variables given in it.
1. Linear Equation Formula
2. Quadratic Equation Formula
3. Cubic Equation Formula
Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x) = ax^{2} + bx + c where a, b, c, ∈ R and a ≠ 0.
1. The roots of the quadratic equation: x = (b ± √D)/2a, where D = b^{2} – 4ac
2. Nature of roots:
 D > 0, roots are real and distinct (unequal)
 D = 0, roots are real and equal (coincident)
 D < 0, roots are imaginary and unequal
3. The roots (α + iβ), (α – iβ) are the conjugate pair of each other.
4. Sum and Product of roots: If α and β are the roots of a quadratic equation, then
 S = α+β= b/a = coefficient of x/coefficient of x^{2}
 P = αβ = c/a = constant term/coefficient of x^{2}
5. Quadratic equation in the form of roots: x^{2} – (α+β)x + (αβ) = 0
6. The quadratic equations a_{1}x_{2} + b_{1}x + c_{1} = 0 and a_{2}x_{2} + b_{2}x + c_{2} = 0 have;
 One common root if (b_{1}c_{2} – b_{2}c_{1})/(c_{1}a_{2} – c_{2}a_{1}) = (c_{1}a_{2} – c_{2}a_{1})/(a_{1}b_{2} – a_{2}b_{1})
 Both roots common if a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}
7. In quadratic equation ax^{2 }+ bx + c = 0 or [(x + b/2a)^{2} – D/4a^{2}]
 If a > 0, minimum value = 4ac – b^{2}/4a at x = b/2a.
 If a < 0, maximum value 4ac – b^{2}/4a at x= b/2a.
8. If α, β, γ are roots of cubic equation ax^{3} + bx^{2} + cx + d = 0, then, α + β + γ = b/a, αβ + βγ + λα = c/a, and αβγ = d/a
9. A quadratic equation becomes an identity (a, b, c = 0) if the equation is satisfied by more than two numbers i.e. having more than two roots or solutions either real or complex.
The solution or roots of a quadratic equation are given by the quadratic formula:
(α, β) = [b ± √(b2 – 4ac)]/2ac
Nature of Roots of Quadratic Equation
Let's see how to determine nature of roots of a quadratic equation through these solved examples:
Example 1: Find the values of k for which the quadratic expression (x – a) (x – 10) + 1 = 0 has integral roots.
 The given equation can be rewritten as, x^{2} – (10 + k)x + 1 + 10k = 0.
D = b^{2} – 4ac = 100 + k^{2} + 20k – 40k = k^{2} – 20k + 96 = (k – 10)^{2} – 4 The quadratic equation will have integral roots, if the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers.
i.e. (k – 10)^{2} – D = 4 Since discriminant is a perfect square. Hence, the difference of two perfect square in R.H.S will be 4 only when D = 0 and (k – 10)^{2} = 4.
⇒ k – 10 = ± 2. Therefore, the values of k = 8 and 12.
Example 2: Find the values of k such that the equation p/(x + r) + q/(x – r) = k/2x has two equal roots.
 The given quadratic equation can be rewritten as:
[2p + 2q – k]x^{2} – 2r[p – q]x + r^{2}k = 0 For equal roots, the discriminant (D) = 0, i.e. b^{2} – 4ac = 0
Here, a = [ 2p + 2q – k ], b = – 2r [ p – q ] and c = r^{2}k
[2r (p – q)]^{2} – 4[(2p + 2q – k) (r^{2}k)] = 0
r^{2}(p – q)^{2} – r^{2}k(2p + 2q – k) = 0 Since r ≠ 0, therefore, (p – q)^{2} – k(2p + 2q – k) = 0
k^{2} – 2(p + q)k + (p – q)^{2}
k = 2(p+q) ± √[4(p + q)^{2} – 4(p – q)]^{2}/2 = (p + q) ± √4pq ∴ The values of k = (p + q) ± 2√pq = (√p ± √q)^{2}
If α and β are roots of a Quadratic Equation ax^{2} + bx + c then:
The relationship between the roots and coefficient of a polynomial equation can be derived by simplifying the given polynomials and substituting the above results as shown below:
Example 3: If the coefficient of x in the quadratic equation x^{2} + bx + c =0 was taken as 17 in place of 13, its roots were found to be 2 and 15. Find the roots of the original quadratic equation.
 Since there is no change in the coefficient of x^{2} and c, therefore, the product of zeros will remain the same for both equations.
 Therefore, the product of zeros (c) = 2 × 15 = 30,
Since, the original value of b is 13. ∴ Sum of zeros = b/a = 13.
 Hence, the original quadratic equation is:
x^{2} – (Sum of Zeros)x + (Product of Zeros) = 0
x^{2} + 13x + 30 = 0
∴ (x + 10) (x + 3) = 0 Therefore, the roots of the original quadratic equations are 3 and 10.
Let's see some solved examples to solve quadratic equations with common roots.
Example 4: For what value of k, both the quadratic equations 6x^{2} – 17x + 12 = 0 and 3x^{2} – 2x + k = 0 will have a common root.
 If one of the root of quadratic Equations a_{1}x^{2} + b_{1}x + c_{1} and a_{2}x^{2} + b_{2}x + c_{2} is common then: (a_{1}b_{2} – a_{2}b_{1})(b_{1}c_{2} – b_{2}c_{1}) = (a_{2}c_{1} – a_{1}c_{2})^{2} . . . . . . . . . . . . . (1)
 Form the given quadratic Equations, a_{1} = 6, b_{1} = 17, c_{1} = 12, a_{2} = 3, b_{2} = 2 and c_{2} = k
 On substituting these values in equation (1), we will get:
[(6×2) – (3×17)] × [17k – (2×12)] = (3×12 – 6k)^{2}
663k + 936 = 1296 + 36k^{2} – 432k
36k^{2} + 231k + 360 = 0
12k^{2} + 125k + 120 = 0
(4k + 15) (3k + 8) = 0 Therefore, the values of k are 154, 83.
Example 5: Find the values of k such that the Quadratic Equations x^{2} – 11x + k and x^{2} – 14x + 2k have a common factor.
 Let (x – α) be the common factor of quadratic Equations x^{2} – 11x + k and x^{2} – 14x + 2k Then x = α will satisfy the given quadratic equations.
 Therefore, α^{2} – 11α + k = 0 . . . . . . . . . . (1)
And, α^{2} – 14α + 2k = 0 . . . . . . . . . . . . (2) On Solving Equation (1) and Equation (2) we will get:
α^{2}/(22k + 14k) = α/2k – k = 1/(14 + 11)
Therefore, α^{2 }= (22k + 14k)/3 = 8/3 . . . . . . . (3)
And, α = (2k – k)/(14 + 11) = k/3 . . . . . . . . . . . (4) On Equating Equation (3) and Equation (4):
8/3 = (k/3)^{2} Therefore, the value of k = 24.
There are two methods to solve a quadratic equation:
General Form: ax^{2} + bx + c = 0;
⇒ x^{2} + bx/a + c/a = 0
⇒ (x + b/2a)^{2} = b^{2}/4a^{2} – c/a
⇒ (x + b/2a)^{2} = (b^{2} – 4ac)/4a^{2}
⇒ x + b/2a = ± (√b^{2} – 4ac)/2a
⇒ x = [b ± √(b^{2} – 4ac)]/2a
b^{2} – 4ac = Discriminant (D)
α = (b+√D)/2a
β = (b – √D)/2a
α+β= b/a, α.β = c/a
Therefore, the quadratic equation can be written as,
⇒ x^{2} – (α + β)x + (α.β) = 0.
Tips to Solve Equations reducible to Quadratic:
 To solve the equations of type ax^{4} + bx^{2} + c = 0, put x^{2} = y
 To solve a.p(x)^{2 }+ b.p(x) + c = 0, put p(x) = y.
 To solve a.p(x) + b/p(x) + c = 0, put p(x) = y.
 To solve a(x^{2} + 1/x^{2}) + b(x + 1/x) + c = 0,put x + 1/x = y and to solve a(x^{2} + 1/x^{2}) + b(x – 1/x) + c = 0, put x – 1/x = y.
 To solve a reciprocal equation of the type ax^{4} + bx^{3} + cx^{2} + bx + a = 0, a ≠ 0, divide the equation by d^{2}y/dx^{2} to obtain a(x^{2 }+ 1/x^{2}) + b(x + 1/x) + c = 0,and then put x + 1/x = y.
 To solve (x + a) (x + b) (x + c) (x + d) + k = 0 where a + b = c + d, put x^{2} +(a + b)x = y
 To solve an equation of type √(ax + b) = cx + d or √(ax^{2} + bx + c) = dx + e, square both the sides.
 To solve √(ax + b) ± √(cx + d) = e, transfer one of the radical to the other side and square both the sides. Keep the expression with radical sign on one side and transfer the remaining expression on the other side.
Graphing Quadratic Equations
The quadratic equation representing a parabola with vertex at P and axis parallel to the yaxis.
Case 1: When a > 0 and b^{2} – 4ac > 0
Case 2: When a > 0 and b^{2} – 4ac = 0
Case 3: When a > 0 and b^{2} – 4ac < 0
Case 4: When a < 0 and b^{2} – 4ac > 0
Case 5: When a < 0 and b^{2} – 4ac = 0
Case 6: When a < 0 and b^{2} – 4ac < 0
Consider a quadratic expression f(x) = ax^{2} + bx + c, where a ≠ 0 and a, b, and c are real. The quadratic expression can be further rewritten as f(x) = x^{2} + bax = ca.
Intervals in which the roots of Quadratic Equation lie
Case 1: Both the roots of a quadratic equation (α, β) are greater than any given number ‘m’ if,
Case 2: Both the roots of a quadratic expression (α, β) are less than any given number ‘m’ if,
Case 3: Both the roots of a quadratic expression (α, β) will lie in the given interval (m1, m2) if,
Case 4: If exactly one root of a quadratic equation (α, β) will lie in the given interval (m1, m2) if, f(m1).f(m2) < 0
Case 5: The given number ‘m’ will lie between the roots of a quadratic Equation α and β if, f(m) < 0
Case 6: The roots of a quadratic equation α and β will have an opposite sign if, f(0) < 0.
Case 7: Both the roots of a quadratic expression α and β are positive if,
Case 8: Both the roots of a quadratic expression α and β are negative if,
Example 6: Find the range of k for which 6 lies between the roots of the quadratic equation x^{2} + 2(k – 3)x + 9 = 0.
 6 will lie between the roots of the quadratic expression f(x) = x^{2 }+ 2(k – 3)x + 9 if,
f (6) < 0
i.e. 36 + 2 (k – 3) . 6 + 9 < 0,
= 36 + 12k – 36 + 9 < 0,
= k < 34 Therefore, the range of k for which 6 lies between the roots of the given quadratic equation is:
k ∈ (∞, 34)
Conditions for Minimum and Maximum Value of Quadratic Equation
How to Find Maximum and Minimum Values of Quadratic Functions?
Case 1
Case 2
Maximum and Minimum Values in Restricted Domain
Case 1
Case 2
Example 7: Find the maximum or minimum value of quadratic equation 4(x – 2)^{2 }+ 2.
 Since the value of ‘a’ is negative, therefore the given quadratic equation will have a maximum value.
 Hence, the maximum value of the quadratic equation 4(x – 2)^{2} + 2 is 2.
Example 8: Find the minimum and maximum values of quadratic equation f(x) = x^{2} – 12x + 11.
 Since a > 0, the maximum and minimum values of a quadratic expression is given by:
[(b^{2 }– 4ac)/4a, ∞) or [D/4a, ∞) Therefore, the minimum value of f(x) is:
(144 – 44)/4 at x = (12/2) = 25 at x = 6. The maximum value of f(x) is infinity. Therefore, the range of the given quadratic equation is [ 25, ∞).
Example 9: How many real roots does the equation 2x^{5} + 2x^{4} – 11x^{3 }+ 9x^{2} – 4x + 2 = 0 will have?
 The given equation has 4 sign changes so it can have a maximum of 4 positive real roots.
 Now for f(x), the equation has only one sign change i.e. f (x) = x^{5} + 2x^{4} + x^{3} + x^{2} + x + 2 = 0. Hence, the equation will have only one negative real root.
Example 10: How many real roots do the quadratic equation x^{2 }+ 3x + 2 will have?
 Since the quadratic equation has no sign change for both f(x) and f(x). Therefore, the equation will have no real root.
 Sign Convention of Quadratic Equation – ax^{2} + bx + c = 0
 The roots of quadratic equation are equal in magnitude but of opposite sign if b = 0 and ac < 0
 The root with greater magnitude is negative if the sign of a = sign of b × sign of c
 If a > 0, c < 0 or a > 0, c > 0; the roots of quadratic equation will have opposite sign
 If y = ax^{2} + bx + c is positive for all real values of x, a > 0 and D < 0
 If y = ax^{2} + bx + c is negative for all real values of x, a < 0 and D < 0
Let's see an example to solve Biquadratic Equations
Biquadratic polynomials can be easily solved by converting them into quadratic equations i.e. by replacing the variable ‘z’ by x^{2}.
Example 11: Find the zeros of a biquadratic equation x^{4 }– 3x^{2} + 2 = 0.
 Given f (x) = x^{4 }– 3x^{2} + 2
 On substituting x^{2} = z in the given equation we get,
f(x) = z^{2} – 3z + 2 = 0
z^{2} – 2z – z + 2 = 0
z(z – 2) 1(z – 2) = 0
∴ z = 1 and z = 2 Hence, x = ±√1 and x = ±√2 [Since, z = x²].
197 videos151 docs200 tests

1. What is a quadratic equation? 
2. What are the roots of a quadratic equation? 
3. How can we determine the nature of the roots of a quadratic equation? 
4. What are the important formulas for solving quadratic equations? 
5. How can we solve quadratic equations? 
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